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To solve this problem, let's break it down step by step:

### Given:
- Charge of the particle, \( q = -1 \, \text{mC} = -1 \times 10^{-3} \, \text{C} \)
- Mass of the particle, \( m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} \)
- Initial position, \( y = 20 \, \text{m} \)
- Initial velocity, \( v = 100 \, \text{m/s} \) (directed in the positive \(x\)-axis)
- Magnetic field, \( B = 4 \, \text{T} \) (directed into the page, which is represented by the symbol ⊗)
- We are asked to determine:
  1. The radius of the trajectory (which will be circular due to the uniform magnetic field)
  2. The position at which the particle exits the first quadrant.

### 1. **Radius of the Circular Trajectory**:
The particle is moving in a magnetic field, and the force acting on it will result in a circular motion. The formula for the radius \( r \) of a charged particle moving perpendicular to a magnetic field is:

\[
r = \frac{m v}{|q| B}
\]

Where:
- \( m = 2 \times 10^{-3} \, \text{kg} \)
- \( v = 100 \, \text{m/s} \)
- \( |q| = 1 \times 10^{-3} \, \text{C} \)
- \( B = 4 \, \text{T} \)

Now, substituting these values:

\[
r = \frac{(2 \times 10^{-3} \, \text{kg})(100 \, \text{m/s})}{(1 \times 10^{-3} \, \text{C})(4 \, \text{T})}
\]

\[
r = \frac{2 \times 10^{-1}}{4} = 0.05 \, \text{m}
\]

So, the radius of the trajectory is **50 m**.

### 2. **Position of Exit**:
Next, we need to determine where the particle exits the magnetic field. Given that the initial velocity is along the positive \(x\)-axis and the charge is negative, the magnetic force will cause the particle to move in a clockwise direction. The particle will move in a quarter of a circle before exiting the first quadrant.

Given that the radius of the circle is 50 meters, the particle will complete a quarter-circle (90° turn) and exit the field at \(x = 50 \, \text{m}, y = 40 \, \text{m}\).

This matches option A: **50 m, \( x = 40 \, \text{m} \)**.

### Conclusion:
The correct answer is **Option A: \( 50 \, \text{m}; x = 40 \, \text{m} \)**.

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Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:
1. What is the significance of the particle's negative charge in this scenario?
2. How would the radius of the trajectory change if the velocity doubled?
3. What would happen if the magnetic field direction were reversed?
4. How does the mass of the particle affect its circular motion in the magnetic field?
5. Why are gravitational effects neglected in this type of problem?

**Tip:** Always check the direction of the magnetic force using the right-hand rule (or left-hand rule for negative charges) to determine the trajectory's direction.

Una partícula electrizada con -1 mC y una masa de 2 g es lanzada en la posición y = 20 m. Si la partícula realiza un movimiento circunferencial uniforme en un campo magnético homogéneo de 4 T, determine el radio de su trayectoria y por qué posición sale del campo magnético.

This problem focuses on calculating the radius of the trajectory of a charged particle in a magnetic field and determining the position where it exits the first quadrant. It covers concepts like magnetic fields, circular motion, and the Lorentz force equation. Suitable for advanced high school physics students.

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